My reasoning for the technique to work for any arbitrary number of mints is that the ratios produced are all unique.
Each mint creates a range of ascending, unique ratios, and no two ranges can overlap.
I am confident that the ranges will never overlap.
However, the ranges aren’t strictly ascending.
This is evident even from the results I posted, starting at just 4 mints.
This raises the possibility that, given enough mints, a ratio may be duplicated within a single range.
I have verified my general construction for up to 25 mints and no duplication has appeared.
But infinity is very large and I must therefore retract my “proof”.

]]>By: David Reynoldshttps://blog.tanyakhovanova.com/2014/06/apsimons-mints/#comment-3122
Thu, 03 Jul 2014 21:16:46 +0000https://blog.tanyakhovanova.com/?p=501#comment-3122Okay. It’s now showing correctly in my browser (Mozilla Firefox).
Things were dropped out when I had less than and greater than symbols in the text.
]]>By: David Reynoldshttps://blog.tanyakhovanova.com/2014/06/apsimons-mints/#comment-3121
Thu, 03 Jul 2014 21:14:54 +0000https://blog.tanyakhovanova.com/?p=501#comment-3121Hmmm… There is a proof here, honest.
I’ll remove all less than and greater than signs and post again.

f(2^n-1) is less than f(2^(n-2))
(2^(n+3)-4n-8)/(3^(n+1)-2n-3) is less than (2^n-2)/(3^(n-1)-1)
(2^(n+3)-4n-8)*(3^(n-1)-1) is less than (2^n-2)*(3^(n+1)-2n-3)
O(2^(n+3)*3^(n-1)) is less than O((2^n)*3^(n+1))
divide both sides by 2^n*3^(n-1)
8 is less than 9

The first assertion holds all n larger than 1.

f(2^(n-1)) is greater than f(2^(n+2)-1)
(2^(n+1)-2)/(3^n-1) is greater than (2^(n+5)-4n-16)/(3^(n+3)-2n-7)
(2^(n+1)-2)*(3^(n+3)-2n-7) is greater than (2^(n+5)-4n-16)*(3^n-1)
O(2^(n+1)*3^(n+3)) is greater than O(2^(n+5)*3^n)
divide both sides by 2^(n+1)*3^n
27 is greater than 16

The second assertion holds for all n larger than 0.

]]>By: David Reynoldshttps://blog.tanyakhovanova.com/2014/06/apsimons-mints/#comment-3120
Thu, 03 Jul 2014 21:12:41 +0000https://blog.tanyakhovanova.com/?p=501#comment-3120I noticed that other statements near the end were also corrupted. Let me try them again.

]]>By: David Reynoldshttps://blog.tanyakhovanova.com/2014/06/apsimons-mints/#comment-3115
Tue, 24 Jun 2014 13:13:21 +0000https://blog.tanyakhovanova.com/?p=501#comment-3115A note concerning the general lower bound.
If r is the number of mints, and r > 1, the number of total coins must be >= ceiling(2^(r/2)).
This is easier to talk about for a discrete case. Let’s take 8 mints as an example.
There are 256 possible outcomes that we must identify with just two weighings, so we must have at least 256 distinct ratios of weight values. For instance, the first weighing might have 20 possible results and the second weighing might only have 13. 20 * 13 > 256. That will work if there aren’t many that reduce to similar ratios. The ideal case, however, would be when the distinct results from both weighing are almost equal. Since the square root of 256 is 16, it is sufficient to have just 16 distinct weight outcomes from each weighing (assuming we can find two sets that have no common ratios). So how many coins do we need to guarantee 16 distinct outcomes?
Let t be the total number of coins.
One outcome is that they are all genuine. weight = t * W(1).
One coin could be fake. weight = (t – 1) * W(1) + W(1 + e).
Two coins could be fake. weight = (t – 2) * W(1) + 2 * W(1 + e).
etc.
We must have at least 15 coins that could be fake to generate 16 distinct outcomes. And this is our minimal answer, 15 total coins.
And now for the second weighing. If we use all the same coins, the ratios will not be unique. If we leave one or more out, we cannot achieve 16 distinct outcomes. Therefore we must add at least 1 more coin to our overall set of coins that we can include in the second weighing. So we must have at least 16 coins, the square root of 2^8.
And in general, that is ceiling(2^(r/2)).
That is more than r, but still very low considering actual results that have been posted. The minimum number of coins to test 6 mints with this formula is 8, but the best case uses 29 coins!
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